A point charge gives rise to an electric field with magnitude 1 N/C at a distance of 3 m. If the distance is increased to 6 m, then what will be the new magnitude of the electric field?

The magnitude of the final electric field is 0.5 N/C.

Explanation:

Given;

initial electric field, E₁ = 1 N/C

initial distance moved by the charge, r₁ = 3 m

final distance moved by the charge, r₂ = 6 m

let the magnitude of the final electric field = E₂

The electric potential created by the charge is given as;

V = E₁r₁ = E₂r₂

Therefore, the magnitude of the final electric field is 0.5 N/C.

divide the right side by the left side

0.024÷1.2=0.02 for example

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge